Cauchy sequences

  • Thread starter steven187
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hello all

I found this rather interesting
suppose that a sequence [tex]{x_{n}}[/tex] satisfies

[tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex] [tex] \forall n\epsilon N[/tex]

how couldnt the sequence [tex]{x_{n}}[/tex] not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

thanxs
 

HallsofIvy

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An obvious one: just add those differences. Let xn be [tex]\sum_{i=1}^n \frac{1}{n}[/tex]. That series does not converge and so the sequence of partial sums is not Cauchy.
 

saltydog

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I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

It looks like a Cauchy sequence to me based on the definition:

The sequence: [itex]\{x_n}\}[/itex] is a Cauchy sequence if given [itex]\epsilon[/itex] there exists N such that for all m,n[itex]\leq[/itex]N we have:

[tex]|x_m-x_n|<\epsilon[/tex]

For the sequence:

[tex] |x_{n+1}-x_{n}|<\frac{1}{n+1}[/tex]


I take [itex]\epsilon=\frac{1}{N+1}[/itex];

Thus for all n>N:

[tex]|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon[/tex]
 
R

rachmaninoff

Thus for all n>N:
[tex]|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon[/tex]
That's true; but look closely - is that really the same thing as

for all m,n[itex]\geq[/itex]N we have:[tex]
|x_m-x_n|<\epsilon[/tex]
?
 

HallsofIvy

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For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
You do remind me that just [tex]|x_{n+1}-x_n|< \epsilon[/tex] is not enough. You have to show that [tex]|x_n-x_m|< \epsilon[/tex] which is basically showing that the tail of the sum goes to 0.
 

saltydog

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rachmaninoff said:
That's true; but look closely - is that really the same thing as



?

I suppose not.

[tex]|x_{n+1}-x_n|<\frac{1}{n+1}[/tex]

does not imply

[tex]|x_m-x_n|<\frac{1}{n+1}[/tex]

Still a little unclear. I'll work on it. Thanks guys.

Seriously, why do I even bother with differential equations anyway . . .
 
R

rachmaninoff

Simply put: for given N, your

[tex] | x_m - x_n | [/tex],

[tex] \mbox{ as } m \rightarrow \infty [/tex],

is not guaranteed to be any better than the (divergent) harmonic series:

[tex] | x_m - x_n | < \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty [/tex]
 

saltydog

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Alright, I'm interested in getting my math right.
Following Hall's example (a slight change):

Let:

[tex]x_n=\sum_{i=1}^{n+1} \frac{1}{i}[/tex]

thus:

[tex]|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}[/tex]

thus:

[tex]|x_{n+1}-x_n|=\frac{1}{n+2}<\frac{1}{n+1}[/tex]

However, the partial sums of this harmonic series [itex] \{x_n\}[/itex] do not converge and thus cannot be a Cauchy sequence.

Now, I got a pot of spagetti to make . . .
 
hello all

nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i cant think of any other examples, do you know of any others?

steven
 
hi guys i need bounded but not cauchy function can you help me asap?
 
Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.
 

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